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XII MATHEMATICS RBSE – FLASHBACK 2017 Time: 3.15 Hrs. Max. Marks: 80

General Instructions:

1. Candidate must write first his / her Roll No. on the question paper compulsorily.

2. All the questions are compulsory.

3. Write the answer to each question in the given answer-book only.

4. For questions having more than one part the answers to those parts are to be written together in

continuity.

5. If there is any error / difference / contradiction in Hindi & English versions of the question paper, the

question of Hindi version should be treated valid.

6. Section Q. Nos. Marks per questions

A 1 – 10 1

B 11 – 25 3

C 26 – 30 5

7. There are internal choices in Q. Nos. 11, 12, 15, 17, 29 and 30. You have to attempt only one of the

alternatives in these questions.

8. Draw the graph of Q. No. 23 on the graph paper

SECTION-A

Q.1 Find the value of 1 1sin sin

3 2

Sol. 1 1sin sin

3 2

1 1sin sin

3 2

sin3 6

sin6

1

2

Q.2 If A = 2 4 1 3

, B3 2 2 5

, then find 2A – B.

Sol. 2A – B = 4 8 1 3 3 5

6 4 2 5 4 1

Q.3 If A = [2 –4 3] and B =

2

4

8

then find (AB)'.

2

Sol. AB = [2 –4 3]

2

4

8

AB = [4 + 16 + 24] = [44]

(AB)' = [44]

Q.4 Find:

21

x dxx

Sol.

21

x dxx

1

x 2 dxx

2

e

xlog x 2x C

2

Q.5 Find the general solution of the differential equation:

2

dy 2x

dx y

Sol. 2

dy 2x

dx y

y2 dy = 2x dx

Integrating on both side, we get

2y dy 2x dx

3

2yx C

3

Q.6 If vector ˆ ˆ ˆa 2i 2j 2k and vector ˆ ˆ ˆb i j k , then find unit vector along the vector (a b) .

Sol. Let ˆ ˆ ˆn a b 3i j k . The unit vector along n is

ˆ ˆ ˆn 3i j k

n̂| n | 11

Q.7 Find the Cartesian form of equation of the line passing through the points (1, 0, 2) and (4, 5, 6)

Sol. Equation of line 1 1 1

2 1 2 1 2 1

x x y y z z

x x y y z z

x 1 y z 2

3 5 4

Q.8 If a line makes 120º, 45º and 90º angles with the x, y and z-axis respectively then find its direction-

cosines.

Sol. We have,

cos = cos 120º = 1

2

o 1m cos cos 45

2

3

on cos cos 90 0

Q.9 Show the region of feasible solution under the following constraints:

x + 3y 6 ; x 0 ; y 0 in answer book.

Sol. x + 3y = 6; x 0, y 0

x 0 6 3

y 2 0 1

(0, 0) does not satisfy x + 3y 6

So, its region lie opposite to the origin

Q.10 If B

P 0.2 and P(A) 0.8,A

then find P(A B).

Sol. B (P B)

PA P(A)

0.2 = (P B)

0.8

P(A B) 0.16

SECTION – B

Q.11 Prove that the relation R defined on set Z as aRB a – b is divisible by 3 is an equivalence relation.

OR

If function f, g : R R are defined as f(x) = x2, g(x) = 2x then find fog(x), gof(x) and fof(3).

Sol. aRb a – b is divisible by 3

(i) Reflexive:

aRa a – a = 0 is divisible by 3

so, R is reflexive

(ii) Symmetric:

Let (a, b) R. Then,

a – b is divisible by 3 a – b = 3 b – a = 3 (–)

b – a is divisible by 3

(b, a) R

so, R is symmetric

(iii) Transitive:

Again (a, b) R and (b, c) R

a – b is divisible by 3 and b – c is divisible by 3 a – b 3 ….(1)

and b – c = 3 ….(2)

4

adding (1) and (2), we get

a – c = 3 ( + )

a – c is divisible by 3

(a, c) R

so, R is transitive. Therefore R is an equivalence reation

OR

f(x) = x2, g(x) = 2x

(i) fog(x) = f[g(x)] = f(2x) = (2x)2 = 4x2

(ii) gof(x) = g[f(x)] = g(x2) = 2x2

(iii) fof(x) = f[f(x)] = f(x2) = (x2)2 = x4

Therefore, fof(3) = (3)4 = 81

Q.12 Express the function tan–1 cos x sin x

cos x sin x

;

x 3x

4 4

in the simplest form.

OR

Prove that: sin–1 8

17 + sin–1 3

5 = tan–1

77

36

Sol. 1 1

sin x1

cos x sin x cos xtan tan

sin xcos x sin x1

cos x

1 1

3x

4 4

1 tan xtan tan tan x x

1 tan x 4 4 4

x2 4 2

x4

OR

L.H.S. 1 1 1 18 3 3sin sin tan tan

17 5 15 4

1 1

1

1

8 3 tan x tan y15 4tan xy 1x y

8 3 tan1 . 1 xy15 4

1 77tan

36

= R.H.S.

Q.13 If 3 1

A1 2

; then prove that A2 – 5A + 7I2 = 0, where I2 is the identity matrix of order 2.

Sol. 3 1

A1 2

5

23 1 3 1

A1 2 1 2

8 5

5 3

Now,

L.H.S. = A2 – 5A + 7I2

8 5 15 5 7 0

5 3 5 10 0 7

0 0

0 0

=R.H.S.

Q.14 Examine the continuity of function f(x) = x 5 x 1

x 5 x 1

at point x = 1.

Sol. f(x) = x 5 , x 1

x 5 , x 1

L.H.L x 1lim f (x)

h 0x 1

lim (x 5) lim(1 h 5) 6

R.H.L x 1lim f (x)

h 0x 1

lim (x 5) lim 1 h 5 4

R.H.L. L.H.L.

So, f(x) is discontinuous at x = 1

Q.15 Find the equation of the tangent to the curve y = x3 – x + 1 at the point whose x coordinate is 1.

OR

The length x of a rectangle is decreasing at the rate 3 cm/minute and the width y is increasing at the

rate 5 cm/minute. When x = 10 cm and y = 6 cm, find the area of the rectangle.

Sol. y = x3 – x + 1 ….(i)

2dy

3x 1dx

at x = 1 , dy

2dx

at x = 1 from the equation of curve y = 1

Therefore, equation of tangent at (1, 1) is

(1,1)

dyy 1 (x 1)

dx

y – 1 = 2(x – 1)

2x – y – 1 = 0

OR

Let at any instant of time t, length be x, breadth y and the area A, then

given that dx

3cm / mindt

dy

5cm / mindt

6

Area A = xy

differentiating with respect to t, we get

dA dy dx

x ydt dt dt

dA

10(5) 6( 3)dt

2dA

32 cm / mindt

Q.16 Find the maximum profit that a company can make, if the profit function is given by:

P(x) = 51 – 72x – 18x2

Sol. We have,

P(x) = 51 – 72x – 18x2

or P'(x) = –72 – 36x

and P''(x) = – 36

Now P'(x) = 0 gives – 72 – 36 x = 0

x = – 2,

at x = –2, P'' (x) < 0

so, P(x) is maximum at x = –2

and P(–2) = 51 + 72 × 2 – 18 × 4

= 51 + 144 – 72

= 123

Q.17 Find: 5

dx

x(x 1)

OR

Find: 1

2

x sin xdx

1 x

Sol. Let 5

6

5

dx dxI

1x(x 1)x 1

x

Put 5 6

1 11 t or ( 5) dx dt

x x

6

1 1dx dt

x 5

Then,

dt 1

I log| t | c5.t 5

5

1 1log|1 | c

5 x

OR

Let 1

2

x sin xI dx

1 x

Put sin–1 x = t

2

1dx dt

1 x

Then,

7

I = t .sin t dt

Integration by parts

t cos t cos t dt

t cos t sin t C

2 1x 1 x sin x C

Q.18 Find: 2

2

sec x dx

tan x 4

Sol. Let 2

2

sec x dxI

tan x 4

Put tan x = t

sec2x dx = dt

Then

2

dtI

t 4

2log t t 4 C

2log tan x tan x 4 C

Q.19 Find the area of the region bounded by parabola y2 = 16x and the lines x = 1, x = 4 and x-axis in the

first quadrant.

Sol.

The required area is

4

1

y.dx

4

1

4 x .dx

43/ 2x

43/ 2

8 56

(8 1) sq.unit3 3

Q.20 Using integration find the area of region bounded by the triangle ABC whose vertices are A(1, 0),

B(2, 2) and C(3, 1)

Sol.

8

Equation of AB

y – 2 = 2 0

(x 2)2 1

y – 2 = 2x – 4

y = 2x – 2

Equation of AC

y – 0 = 1 0

(x 1)3 1

1

y (x 1)2

Equation of BC

y – 2 = 2 1

(x 2)2 3

y – 2 = – x + 2 y = –x + 4

Required area = area of ABD + area of DBCE – area of ACE

2 3 3

1 2 1

x 1(2x 2)dx (4 x)dx dx

2

3 22 2

2 2

1

2 1

x 1 x(x 2x) 4x x

2 2 2

= 9 1 9 1

[(4 4) (1 2)] 12 (8 2) 3 12 2 2 2

3 1

1 (2)2 2

3

2

Q.21 If a 5i j 3k and b i 3j 5k , then find the angle between the vectors (a b) and (a b) .

Sol. a b 6i 2j 8k

a b 4i 4j 2k

If is angle between (a b) and (a b) . Then

(a b).(a b)

cosa b a b

or 24 8 16

cos 0104 36

or 90o

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Q.22 Find the area of a parallelogram whose adjacent sides are vectors a i j 3k and

b 2i 7j k .

Sol. We have,

i j k

a b 1 1 3

2 7 1

Or a b i( 1 21) j(1 6) k( 7 2)= ˆ ˆ ˆ20i 5j 5k

The required area is

= a b 400 25 25 450 15 2

Q.23 By graphical method solve the following linear programming problem for minimize.

Objective function Z = 5x + 7y

Constraints 2x + y ≥ 8

x + 2y ≥ 10

x ≥ 0, y ≥ 0

Sol. Z = 5x + 7y

2x + y ≥ 8

x + 2y ≥ 10

x ≥ 0, y ≥ 0

Solving 2x + y = 8 and x + 2y = 10

(x, y) = (2, 4)

Corner Point Value of Z

(10, 0) 50

(0, 8) 56

(2,4) 38

Since, region is unbounded. Consider z < 38 or 5x + 7y < 38

Clearly. It has no points in common with feasible region.

Thus, the minimum value of Z is 38 at (2, 4).

10

Q.24 Given three identical boxes I, II and III each containing two coins. In box I both coins are gold coins

in box II both are silver coins and in the box III there is one gold and one silver coin. A person

chooses a box at random and take out a coin. If the coin is of silver what is the probability that the

other coin in the box is also of silver.

Sol. Let E1, E2 and E3 be the events that boxes I, II and III are chosen, respectively. Then

P(E1) = P(E2) = P(E3) 1

3

Also, let A be the event that ‘Coin drawn is of silver’.

Then, 1

AP 0

E

2

AP 1

E

3

A 1P

E 2

The required probability is

2

22

1 2 3

1 2 3

AP(E ).P

EEP

A A A AP(E ).P P(E ).P P(E ).P

E E E

11

1 231 1 1 1 3/ 2 3

0 13 3 3 2

Q.25 Find the variance of the number obtained on a throw of an unbiased die.

Sol. Sample space S = {1, 2, 3, 4, 5, 6}

let x denote number oftained on the throw. Then x = 1,2,3,4,5 or 6

also, P(1) = P(2) = P(3) = P(4) = P(5) = P(6) =1

6

Therefore, the probability distribution

x 1 2 3 4 5 6

P(x) 1

6

1

6

1

6

1

6

1

6

1

6

Now, n

1

E(x)= ( )i i

i

x P x

= 1 1 1 1 1 1

1 2 3 4 5 66 6 6 6 6 6

= 21

6

and 2 2 2 2 2 21 1 1 1 1

( ) 1 2 3 5 66 6 6 6 6

E x

= 91

6

11

Therefor, Var (x) = 2 2( ) ( ( ))E x E x

291 21

6 6

35

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SECTION-C

Q.26 Show that

2 3

2 3

2 3

a a 1 pa

b b 1 pb

c c 1 pc

= (1 + pabc) (a – b) (b – c) (c – a)

Sol. Let

2 3

2 3

2 3

a a 1 pa

b b 1 pb

c c 1 pc

2 2 3

2 2 3

2 2 3

a a 1 a a pa

b b 1 b b pb

c c 1 c c pc

2 2

2 2

2 2

a a 1 1 a a

b b 1 abcp 1 b b

c c 1 1 c c

= (1 + pabc)

2

2

2

a a 1

b b 1

c c 1

Apply operation 2 2 1 3 3 1R R R , R R R

2

2 2

2 2

a a 1

(1 abcp) b a b a 0

c a c a 0

=

2a a 1

(1 abcp) (b a) (c a) 1 b a 0

1 c a 0

= 1 b a

(1 abcp)(b a)(c a)1 c a

= (1 + abcp) (b – a) (c – a) (c + a – b – a)

= (1 + abcp) (a – b) (b – c) (c – a)

Q.27 If y = x2 + xp + px + pp > 0 and x > 0, then find dy

dx

Sol. y = xx + xp + px + pp. differentiating with respect to x, we get

x p 1 x

e

dy d(x ) px p log p 0

dx dx

x p 1 x

e

d(x ) px p log p

dx

Now, consider, z = xx

or log z = log xx = x log x

12

differentiating with respect to x,

we get

1 dz 1

x. log xz dx x

dz

z (1 log x)dx

x xd

(x ) x (1 log x)dx

Therefore,

x p 1 x

e

dyx (1 log x) px p .log p

dx

Q.28 Show that 2

2 2 2 2

0

x dx

a cos x b sin x 2ab

Sol. Let

2 2 2 2

0

x dxI

a cos x b sin x

….(1)

or 2 2 2 2

0

( x)dxI

a cos ( x) b sin ( x)

a a

0 0

f (x)dx f (a x)dx

2 2 2 2

0

( x)dxI

a cos x b sin x

….(2)

Adding (1) & (2), we get

2 2 2 2

0

dx2I

a cos x b sin x

/ 2

2 2 2 2

0

dx2I 2

a cos x b sin x

[ f(a – x) = f(x)]

/ 2 2

2 2 2

0

sec x dxI

a b tan x

[divided by cos2 x]

put 2tan x t sec x dx dt

2 2 2

0

dtI

a b t

22

202

dtI

abt

b

1 1 1

2

0

1 tI . tan tan tan 0

a ab ab

b b

13

I . 0ab 2

I2ab

Proved

Q.29 Find the solution of the differential equation (x – y) dy – (x + y) dx = 0.

OR

Find the solution of the differential equation

2 dycos x. y tan x 0 x

dx 2

Sol. (x – y) dy – (x + y) dx = 0

dy x y

dx x y

put y = vx

dy dv

v xdx dx

dv x vx

v xdx x vx

dv 1 v

x vdx 1 v

2dv 1 v v v

xdx 1 v

2

1 v dxdv

1 v x

2 2

1 v dxdv dv

1 v 1 v x

tan–1v – 1

2 log (1 + v2) = logx + C

2

1

2

y 1 ytan log 1 log x C

x 2 x

1 2 2y 1tan log x y C

x 2

OR

2 dy

cos x . y tan xdx

2 2

dy y tan x

dx cos x cos x

Which is linear equation in ‘y’

Here, P = 2 2

2

1sec x,Q sec x .tan x

cos x

Now, I.F. = 2P.dx sec xdx tan xe e e

Therefore, solution is given by

tan x tan x 2ye e .tan x.sec x dx c

Put tan x = z

sec2x dx = dz. Then,

14

tan x z z z z z zye e .z dz c ze 1.e dz c ze e c (z 1)e c

tan x tan xye (tan x 1)e c

y = (tan x – 1) + ce–tan x

Q.30 Find the shortest distance between the lines:

ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆr (i 2j 3k) ( i j 2k) and r (i j k) (i 2j 2k)

OR

Find the equation of the plane that contains the point (2, –1, 3) and is perpendicular to each of the

planes 2x + 3y – 2z = 5 and x + 2y – 3z = 8

Sol. Here,

i 2 1 2

ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆa i 2j 3k; a i j k;b i j 2k; b i 2j 2k

Shortest distance = 2 1 1 2

1 2

(a a ).(b b )

b b

Now,

1 2

ˆ ˆ ˆi j k

b b 1 1 2

1 2 2

ˆ ˆ ˆi( 2 4) j(2 2) k( 2 1)

ˆ ˆ ˆ2i 4j 3k

and 2 1

ˆ ˆa a j 4k

Therefore, the shortest distance is

( 4 12) 8

4 16 9 29

OR

Let the equation of plane

a(x – 2) + b(y + 1) + c(z – 3) = 0 ….(1)

Plane (1) is perpendicular to the planes

2x + 3y – 2z = 5 and x + 2y – 3z = 8

Then,

2a + 3b – 2c = 0 ….(2)

a + 2b – 3c = 0 ….(3)

Solving (2) & (3)

a b c

5 4 1

From (1), we get

–5(x – 2) + 4(y + 1) + (z – 3) = 0

–5x + 4y + z + 11 = 0

5x – 4y – z – 11 = 0